Verified Solution • 2025–26 Session moderate

QJEEM18Ph6590: JEE Main 2018 Physics - Class 12 JEEM18Ph6590 Solution: An ideal capacitor of capacitance is charged to a potential…

An ideal capacitor of capacitance 0.2 μF0.2\,\mu F is charged to a potential difference of 1010 V.V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.50.5 mH.mH. The current at a time when the potential difference across the capacitor is 55 V,V, is :

Multiple Choice Options

  • A 0.34  A0.34\,\,A
  • B 0.25  A0.25\,\,A
  • C 0.17  A0.17\,\,A
  • D 0.15  A0.15\,\,A

Step-by-Step Verified Solution

Capacitance, C = 0.2 μ\mu F = 0.2 ×\times 10 −- 6 F

Inductance, L = 0.5 m H = 0.5 ×\times 10 −- 3 H

Let, current = I.

Using energy conservation,

UE + 0 = UE' + Ub'

⇒\Rightarrow     \,\,\,\, 12{1 \over 2} cv2 + 0 = 12{1 \over 2} c v12v_1^2 + 12{1 \over 2} LI2

⇒\Rightarrow     \,\,\,\, 12{1 \over 2} ×\times 0.2 ×\times 10 −- 6 ×\times 102

= 12×{1 \over 2} \times 0.2 ×\times 10 −- 6 ×\times 52 + 12{1 \over 2} ×\times 0.5 ×\times 10 −- 3 ×\times I2

By solving this,

I = 3×\sqrt 3 \times 10 −- 1 A

= 0.17 A.

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